IIT JEE , AIEEE Forum - Mathematics , Algebra

Werewolf

Sum up the series :

S = 1/2 + 2/3 + 3/4 + .........n/(n+1)

sandeepramesh

i dont think this may be possible bcos it reduces to n - (1/2 + 1/3.............) which amounts to finding the sum of an Hp of n terms which hasnt been done till now

Sushmi

Can't we proceed using binomial theorem for the later part

(1 - x) ^ n = C0 - C1x + C2 x^2 - C3 x^3 ......+(-1)^n x^n

so we can write it like,

1 - (1-x)^n / x = C1 - C2 x^2 + C3 x^3 - ...+( -1)Cn x^n-1

or _[0]

^[1] 1 - (1- x)^n / 1 - (1-x) = [ C1x + C2 x^2/2 - C3 x^3/3 ......+(-1)^n x^n/n ]

wont this give the required value for the later part?

sandeepramesh

no it wont as the nr is 1 for all :)

Sushmi

Yes but we need numerator like 1/2,1/3,1/4......then?please be a little more clear.

sandeepramesh

NO I MEANT:

The series sum = n - (1/2 + 1/3 + 1/4 + 1/5........)

You cannot get 1 in all the nrs by using binomial methods as Cr not = 1 for all r :)

computer001

well actually tht is equal to 1 - log0 so v cant do nethin wid it :\
log(1+x) = x -x^2/2 + x^3/3...
as x->-1 this tends to -1 -1/2-1/3-1/4...

Sushmi

_[0]

^[1] 1 - (1- x)^n / 1 - (1-x) = [ C1x + C2 x^2/2 - C3 x^3/3 ......+(-1)^n x^n/n ]

= _[0]

^[1] {1 + (1-x) + (1-x) ^2 + ......+(1-x)^n-1 }dx

= [ x + (1-x)^2/-2 + ( 1-x)^3/-3 +.....+(1-x)^n/-n ] within limits of 0 to 1

= 1 + 1/2 + 1/3 + ......+1/n

= C1 - 1/2 C2 + 1/3 C3 - .....+(-1) ^ n-1 1/n Cn

We need the L.H.S only.

So now the problem becomes = n -[C1 - 1/2 C2 + 1/3 C3 - .....+(-1) ^n-1 1/n Cn]

Isn't it? I was speaking of this.

sandeepramesh

who said 1/r * Cr = 1???

as to computer001 's method, unfortunately, log(1+x) is an infinite series while we want sum to n terms :)

konichiwa2x

i havent inspected this closely, but i dont think this can be done. sandeep is right. you wont be able to get rid of the nCr terms..

computer001

yeah i realised tht but u cud consider n is suff large n after tht it becomes practically negligible..

Werewolf

Oh wow ! Heavy discussion going on.

Guys let go off this question, no point wasting time , I have asked a dozen people but none came up with a positive answer, so I think this series cannot be summed up.

Sushmi

I never said 1/r Cr = 1

.

Conjurer

S = 1/2 + 2/3 + 3/4 + .........n/(n+1)

Write S as (1-1/2) + (1-1/3) + (1-1/4)...... 1-1/(n+1)
S = n -(1/2 + 1/3 + 1/4 + ....1/n+1)

Now can anyone tell me how to proceed further?

EDIT: Oops sandeep had posted it already.

Ask & Discuss Questions with Community & Experts